Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 + 2}{x - 4} = \dfrac{6x - 6}{x - 4}$
Solution: Multiply both sides by $x - 4$ $ \dfrac{x^2 + 2}{x - 4} (x - 4) = \dfrac{6x - 6}{x - 4} (x - 4)$ $ x^2 + 2 = 6x - 6$ Subtract $6x - 6$ from both sides: $ x^2 + 2 - (6x - 6) = 6x - 6 - (6x - 6)$ $ x^2 + 2 - 6x + 6 = 0$ $ x^2 + 8 - 6x = 0$ Factor the expression: $ (x - 4)(x - 2) = 0$ Therefore $x = 4$ or $x = 2$ However, the original expression is undefined when $x = 4$. Therefore, the only solution is $x = 2$.